Chapter 2.3, Problem 98E

To determine

To find the real zeroes of the polynomial function $h(x)=x^5+5x^4-5x^3-15x^2-6x$ .

Answer to Problem 98E

  $-1,0,2,-3+\sqrt{6},-3-\sqrt{6}$

Explanation of Solution

Given:

Function: $h(x)=x^5+5x^4-5x^3-15x^2-6x$

Calculation:

The leading coefficient of given polynomial is 1 and the constant term is 0.

So, possible zeroes are: $\frac{\text{Factors of 0}}{\text{Factors of 1}}=\frac{\pm \text{0}}{\pm 1}=0$

Using synthetic division to check whether x = 0, is a zero or not.

  $\begin{array}{ccccccc}0& 1& 5& -5& -15& -6& 0\\ & & 0& 0& 0& 0& 0\\ & 1& 5& -5& -15& -6& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, x = 0 is a zero of the given polynomial.

So, $h(x)$ factors as

  $h(x)=(x-0)(x^4+5x^3-5x^2-15x-6)$

Now, consider $x^4+5x^3-5x^2-15x-6.$

The leading coefficient of above polynomial is 1 and the constant term is -6.

So, possible zeroes are: $\frac{\text{Factors of}-6}{\text{Factors of 1}}=\frac{\pm 1,\pm 2,\pm 3,\pm 6}{\pm 1}=\pm 1,\pm 2,\pm 3,\pm 6$

Using synthetic division to check whether x = 1, is a zero or not.

  $\begin{array}{cccccc}1& 1& 5& -5& -15& -6\\ & & 1& 6& 1& -14\\ & 1& 6& 1& -14& \boxed{-22}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, remainder is not 0. So, x = 1 is not a zero of the given polynomial.

Using synthetic division to check whether x = 2, is a zero or not.

  $\begin{array}{cccccc}2& 1& 5& -5& -15& -6\\ & & 2& 14& 18& 6\\ & 1& 7& 9& 3& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, x = 2 is a zero of the given polynomial.

So, $h(x)$ factors as

  $h(x)=(x-0)(x-2)(x^3+7x^2+9x+3)$

Now, consider $x^3+7x^2+9x+3.$

The leading coefficient of above polynomial is 1 and the constant term is 3.

So, possible zeroes are: $\frac{\text{Factors of 3}}{\text{Factors of 1}}=\frac{\pm 1,\pm 3}{\pm 1}=\pm 1,\pm 3$

Using synthetic division to check whether x = -1, is a zero or not.

  $\begin{array}{ccccc}-1& 1& 7& 9& 3\\ & & -1& -6& -3\\ & 1& 6& 3& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, x = -1 is a zero of the given polynomial.

So, $h(x)$ factors as

  $h(x)=(x-0)(x-2)(x+1)(x^2+6x+3)$

Now, consider $x^2+6x+3.$

The above polynomial is a second degree polynomial.

So, zeroes of a second degree polynomial can be found using the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$

  $\begin{array}{l}\Rightarrow x=\frac{-(6)\pm \sqrt{{(6)}^2-4(1)(3)}}{2(1)}\\ \Rightarrow x=\frac{-6\pm \sqrt{36-12}}{2}\\ \Rightarrow x=\frac{-6\pm \sqrt{24}}{2}\\ \Rightarrow x=\frac{-6\pm 2\sqrt{6}}{2}\\ \Rightarrow x=-3\pm \sqrt{6}\end{array}$

Conclusion:

Therefore, the zeroes of given polynomial function are $x=-1,0,2,-3+\sqrt{6}\text{and}-3-\sqrt{6}.$